100 ml of 0.01 M `H_2 SO_4` is titrated against 0.2 M `Ca (OH)_2`, volume of `Ca (OH)_2` required to reach end point will be :

Options

A. 5 ml
B. 10 ml
C. 20 ml
D. 15 ml

Answer

To solve the problem of titrating 100 ml of 0.01 M `H2SO4` against 0.2 M `Ca(OH)2`, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical reaction between sulfuric acid (`H2SO4`) and calcium hydroxide (`Ca(OH)2`) is:

\[ H_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 + 2H_2O \]

From the equation, we can see that 1 mole of `H2SO4` reacts with 1 mole of `Ca(OH)2`.

### Step 2: Calculate the moles of `H2SO4`
To find the number of moles of `H2SO4`, we use the formula:

\[ \text{Moles} = \text{Molarity} \times \text{Volume} \]

Given:
- Molarity of `H2SO4` = 0.01 M
- Volume of `H2SO4` = 100 ml = 0.1 L (since 1 L = 1000 ml)

Calculating the moles:

\[ \text{Moles of } H_2SO_4 = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{mol} \]

### Step 3: Determine the moles of `Ca(OH)2` required
From the balanced equation, we know that 1 mole of `H2SO4` reacts with 1 mole of `Ca(OH)2`. Therefore, the moles of `Ca(OH)2` required will also be:

\[ \text{Moles of } Ca(OH)_2 = 0.001 \, \text{mol} \]

### Step 4: Calculate the volume of `Ca(OH)2` required
Using the formula for moles again, we can rearrange it to find the volume:

\[ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} \]

Given:
- Molarity of `Ca(OH)2` = 0.2 M

Calculating the volume:

\[ \text{Volume of } Ca(OH)_2 = \frac{0.001 \, \text{mol}}{0.2 \, \text{mol/L}} = 0.005 \, \text{L} \]

To convert this to milliliters:

\[ 0.005 \, \text{L} = 5 \, \text{ml} \]

### Final Answer
The volume of `Ca(OH)2` required to reach the endpoint is **5 ml**.

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